Question: Simplify the following expression and state the condition under which the simplification is valid. $r = \dfrac{y^2 - 16}{y - 4}$
Answer: First factor the polynomial in the numerator. The numerator is in the form ${a^2} - {b^2}$ , which is a difference of two squares so we can factor it as $({a} + {b})({a} - {b})$ $ a = y$ $ b = \sqrt{16} = -4$ So we can rewrite the expression as: $r = \dfrac{({y} {-4})({y} + {4})} {y - 4} $ We can divide the numerator and denominator by $(y - 4)$ on condition that $y \neq 4$ Therefore $r = y + 4; y \neq 4$